Question

Problem 5: Select the lightest steel beam. Fy=50 ksi. The lateral bracing is provided at the both ends and at the two points 10 ft each from the ends. Neglect Mn Fy 2 weight of the beam. (1) Check moment. (2) Check shear. (3) Determine the maximum deflection. 50 Pin Pin Po=22k PL=45k PD-22k PL 45k toft 10'x3=30'

Answer 1

(1)

Step1 :

D.L=22 kips

L.L=45 kips

Pu=1.2(22)+1.6(45)

Pu=98.4 kips is the point load acting on each side

Step 2:

Maximum bending moment=WL/3

B.M=98.4x30/3

B.M=984 ft-kips

Step 3:

Calculate the section modulus

Zx=Mu/(0.9Fy)

Zx=984x12/(0.9x50)

Zx=262.4 in³

From the Zx tables, select a W section and consider self weight also

Step 4:

Consider W21x111 having Zx=279 in³,

Now check for self weight

D.L=1.2(22+0.111)

L.L=1.6(45)

Pu=98.53 kips

M=98.51x10=985.3 ft-kips

Zx=Mu/(0.9Fy)

Zx=985.3x12/(0.9x50)

Zx=263 in³ is required, 279 in³ is provided

Step 5:

Check for Lp, Lb and Lr values

Lb=10 ft

Lp=10.2 ft

Lr=31.3 ft

As Lp>Lb, no local torsional buckling occurs

Mn=Mp=FyZx

Mp=50x279

Mp=13950 kip-in=1162.5 ft-kips

LRFD design strength=0.9x1162.5=1046.25 ft-kips

$\phi$Mn>Mu, so the beam is safe in bending

(2)

Check for shear:

Vu=98.4 kips

Vn=0.6Fydtw

Vn=0.6x50x21.5x0.55

Vn=354.75 kips, this beam is safe in shear also

(3)

P=45 kips (deflection is calculated only for unfactored live load)

Span/240=30x12/240=1.5 in

$\delta$=Pa(3L²-4a²)/24EI

$\delta$=45x120(3x360²-4x120²)/24x29000x2670

$\delta$=0.96 in

The beam W21X111 is safe in bending, shear and deflection, so it can be safely used